integral of (6x^2)/((x^6+1)(x^3+1))

Lösung

\int \frac{6 x ^{2}}{( x ^{6} + 1 ) ( x ^{3} + 1 )} d x

- \frac{1}{2} ln x^{6} + 1 + arctan (x^{3}) + ln x^{3} + 1 + C

Schritte zur Lösung

\int \frac{6 x ^{2}}{( x ^{6} + 1 ) ( x ^{3} + 1 )} d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int \frac{x ^{2}}{( x ^{6} + 1 ) ( x ^{3} + 1 )} d x

Wende U-Substitution an

= 6 \cdot \int \frac{1}{3 ( u ^{2} + 1 ) ( u + 1 )} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \frac{1}{3} \cdot \int \frac{1}{( u ^{2} + 1 ) ( u + 1 )} d u

Ermittle den Partialbruch von

\frac{1}{( u ^{2} + 1 ) ( u + 1 )} : \frac{- u + 1}{2 ( u ^{2} + 1 )} + \frac{1}{2 ( u + 1 )}

= 6 \cdot \frac{1}{3} \cdot \int \frac{- u + 1}{2 ( u ^{2} + 1 )} + \frac{1}{2 ( u + 1 )} d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 6 \cdot \frac{1}{3} (\int \frac{- u + 1}{2 ( u ^{2} + 1 )} d u + \int \frac{1}{2 ( u + 1 )} d u)

\int \frac{- u + 1}{2 ( u ^{2} + 1 )} d u = \frac{1}{2} (- \frac{1}{2} ln u^{2} + 1 + arctan (u))

\int \frac{1}{2 ( u + 1 )} d u = \frac{1}{2} ln ∣ u + 1 ∣

= 6 \cdot \frac{1}{3} (\frac{1}{2} (- \frac{1}{2} ln u^{2} + 1 + arctan (u)) + \frac{1}{2} ln ∣ u + 1 ∣)

Setze in

u = x^{3}

ein

= 6 \cdot \frac{1}{3} (\frac{1}{2} (- \frac{1}{2} ln (x^{3})^{2} + 1 + arctan (x^{3})) + \frac{1}{2} ln x^{3} + 1)

Vereinfache

6 \cdot \frac{1}{3} (\frac{1}{2} (- \frac{1}{2} ln (x^{3})^{2} + 1 + arctan (x^{3})) + \frac{1}{2} ln x^{3} + 1) : - \frac{1}{2} ln x^{6} + 1 + arctan (x^{3}) + ln x^{3} + 1

= - \frac{1}{2} ln x^{6} + 1 + arctan (x^{3}) + ln x^{3} + 1

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{2} ln x^{6} + 1 + arctan (x^{3}) + ln x^{3} + 1 + C